从3个表格中选择PHP

Question

起初，我为我的英语不好而感到抱歉，但是我需要用GET方法做一个表格来选择3个表格的数据

where $_GET['steamid']: 
achievements (id, name, description);
progress (id_player, achieve_id, progress);
global (id_player, steamid, nick);

to put in a table, example:
Achievement | Progress 
name_achievement | 5

就像在： http : //achievements.my-run.de/index.php?server=jail

Answer 1

我听不懂您的意思，但也许是这样？

SELECT achievements.id, achievements.name, achievements.description, progress.id_player, progress.archieve_id, progress.progress, global.id_player, global.steamid, global.nick
FROM global INNER JOIN achievements ON global.id_player = achievements.id INNER JOIN progress ON progress.id_player = achievements.id
WHERE global.steamid = STEAMID

Answer 2

你需要使用一个join加入你的表到一个查询。 我认为这应该有效，只需绑定GET值。

select a.name, p.progress from global as g
join progress as p
on g.id_player = p.id_player
join achievements as a
on p.achieve_id = a.id
where g.steamid = ?

Answer 3

多亏了所有人，我已经为乱七八糟和我的英语不好而感到抱歉

我做的：

$sql_global = "SELECT Id FROM promad_mglobal WHERE SteamId='".$steam."'";
    $result_global = $con->query($sql_global);
    while($row_global = $result_global->fetch_assoc())
        $idd = $row_global['Id'];

    $sql_medals = "SELECT * FROM promad_medalhas 
    LEFT JOIN promad_mprogresso ON (promad_mprogresso.Achievement = promad_medalhas.Id AND promad_mprogresso.Id='$idd')";

    $result_medals = $con->query($sql_medals);
    while($row_medals = $result_medals->fetch_assoc())
    {       
        if(empty($row_medals['Progress']))
            $row_medals['Progress'] = 0;

        echo "- ".$row_medals['Name']." ".$row_medals['Progress']." <br>";
    }

而且有效！

谢谢